What is the ph of a solution of 0.400 m ch₃nh₂ containing 0.250 m ch₃nh₃i? (kb of ch₃nh₂ is 4.4 × 10⁻⁴)

The pH of a solution of 0.400 M CH₃NH₂ containing 0.250 M CH₃NH₃I = 10.84

This given solution contains a weak base (CH₃NH₂) and the conjugate acid of that weak base (CH₃NH₃I) which makes a Buffer.

We know that  

or,

CH₃NH₂ is the base and CH₃NH₃I is the corresponding conjugate acid (salt).

Given:

[base] = 0.400 M

[salt] or [acid] = 0.250 M

= 4.4 * 10-4

So = 3.36

solution:

Putting in the equation,

pOH= 3.16

So pH = 14 - pOH

= 10.84

Thus, the pH of a solution of 0.400 M CH₃NH₂ containing 0.250 M CH₃NH₃I = 10.84

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Explanation:

A solution of a weak base and its conjugate acid is a buffer.

The equation for the equilibrium is

The Henderson-Hasselbalch equation for a basic buffer is

Data:

   [B] = 0.400 mol·L⁻¹

[BH⁺] = 0.250 mol·L⁻¹

    Kb = 4.4 × 10⁻⁴

Calculations:

(a) Calculate pKb

pKb = -log(4.4× 10⁻⁴)  = 3.36

(b) Calculate the pH

they both have a y in them

the right answer is c i just did the question.